Area of Parallelogram

A line '$L$' is drawn from $P\left(4,3\right)$ to meet the lines $L_1:3x+4y+5=0$ and $L_2:3x+4y+15=0$ at point $A$ and $B$ respectively. From '$A$' a line, perpendicular to $L$ is drawn meeting the line $L_2$ at $A_1.$ Similarly from point '$B$' a line, perpendicular to $L$ is drawn meeting the line $L_1$ at $B_1.$ Thus a parallelogram $A{A}_{1}B{B}_1$ is formed. Find the equation(s) of '$L$' so that the area of the parallelogram $A{A}_{1}B{B}_1$ is least.

Prove that the area of the parallelogram formed by the lines $3x-4y+a=0, 3x-4y+3a=0, 4x-3y-a=0$ and $4x-3y-2a=0$ is $\frac{2a^2}{7}$ sq. units